Extended Kalman Filter measurement residual computation
Multi tool use
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I am trying to follow the computation of EKF presented in this paper http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=C9CB210A45F0D7ED5CA7DE174F1A5490?doi=10.1.1.681.8390&rep=rep1&type=pdf (p.16-18), with the only difference that my system state does not include the acceleration, so it is a $4x1$ matrix:
$$x_x(t) = begin{bmatrix}
x(t) \
v_x(t) \
y(t) \
v_y(t)
end{bmatrix} $$
where $x(t)$ and $y(t)$ are the cartesian coordinates and $v_x(t)$ and $v_y(t)$ are the velocity components.
I have problem computing the State Estimate Update step (eq. 17 in the paper), where the measurement residual needs to be calculated. According to the paper, the State Estimate Update is:
$$hat{x}_x(k^{+}) = hat{x}_x(k^{-}) + K(k)[z_{z}(k)-h(hat{x}_x(k^{-}))]$$
where
$$z_z(k) = begin{bmatrix}
z_x(k) \
z_y(k)
end{bmatrix},
$$
where $z_x(k)$ and $z_y(k)$ are the measurements of $x$ and $y$ positions and
$$
h(x_x(k)) = begin{bmatrix}
x(k) &0 &0 &0 \
0 &0 &y(k) &0
end{bmatrix}. $$
$K(k)$ is the Kalman Gain, which in my case is a $4x2$ matrix.
My question is, how do I subtract $h(hat{x}_x(k^{-}))$ which is a $2x4$ matrix from $z_z(k)$, which is a $2x1$ matrix?
If I am not mistaken, the result of $K(k)[z_{z}(k)-h(hat{x}_x(k^{-}))]$ needs to be a $4x1$ matrix so that I can later add it to the previous estimate $hat{x}_x(k^{-})$, which is also $4x1$.
kalman-filter
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I am trying to follow the computation of EKF presented in this paper http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=C9CB210A45F0D7ED5CA7DE174F1A5490?doi=10.1.1.681.8390&rep=rep1&type=pdf (p.16-18), with the only difference that my system state does not include the acceleration, so it is a $4x1$ matrix:
$$x_x(t) = begin{bmatrix}
x(t) \
v_x(t) \
y(t) \
v_y(t)
end{bmatrix} $$
where $x(t)$ and $y(t)$ are the cartesian coordinates and $v_x(t)$ and $v_y(t)$ are the velocity components.
I have problem computing the State Estimate Update step (eq. 17 in the paper), where the measurement residual needs to be calculated. According to the paper, the State Estimate Update is:
$$hat{x}_x(k^{+}) = hat{x}_x(k^{-}) + K(k)[z_{z}(k)-h(hat{x}_x(k^{-}))]$$
where
$$z_z(k) = begin{bmatrix}
z_x(k) \
z_y(k)
end{bmatrix},
$$
where $z_x(k)$ and $z_y(k)$ are the measurements of $x$ and $y$ positions and
$$
h(x_x(k)) = begin{bmatrix}
x(k) &0 &0 &0 \
0 &0 &y(k) &0
end{bmatrix}. $$
$K(k)$ is the Kalman Gain, which in my case is a $4x2$ matrix.
My question is, how do I subtract $h(hat{x}_x(k^{-}))$ which is a $2x4$ matrix from $z_z(k)$, which is a $2x1$ matrix?
If I am not mistaken, the result of $K(k)[z_{z}(k)-h(hat{x}_x(k^{-}))]$ needs to be a $4x1$ matrix so that I can later add it to the previous estimate $hat{x}_x(k^{-})$, which is also $4x1$.
kalman-filter
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0
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up vote
0
down vote
favorite
I am trying to follow the computation of EKF presented in this paper http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=C9CB210A45F0D7ED5CA7DE174F1A5490?doi=10.1.1.681.8390&rep=rep1&type=pdf (p.16-18), with the only difference that my system state does not include the acceleration, so it is a $4x1$ matrix:
$$x_x(t) = begin{bmatrix}
x(t) \
v_x(t) \
y(t) \
v_y(t)
end{bmatrix} $$
where $x(t)$ and $y(t)$ are the cartesian coordinates and $v_x(t)$ and $v_y(t)$ are the velocity components.
I have problem computing the State Estimate Update step (eq. 17 in the paper), where the measurement residual needs to be calculated. According to the paper, the State Estimate Update is:
$$hat{x}_x(k^{+}) = hat{x}_x(k^{-}) + K(k)[z_{z}(k)-h(hat{x}_x(k^{-}))]$$
where
$$z_z(k) = begin{bmatrix}
z_x(k) \
z_y(k)
end{bmatrix},
$$
where $z_x(k)$ and $z_y(k)$ are the measurements of $x$ and $y$ positions and
$$
h(x_x(k)) = begin{bmatrix}
x(k) &0 &0 &0 \
0 &0 &y(k) &0
end{bmatrix}. $$
$K(k)$ is the Kalman Gain, which in my case is a $4x2$ matrix.
My question is, how do I subtract $h(hat{x}_x(k^{-}))$ which is a $2x4$ matrix from $z_z(k)$, which is a $2x1$ matrix?
If I am not mistaken, the result of $K(k)[z_{z}(k)-h(hat{x}_x(k^{-}))]$ needs to be a $4x1$ matrix so that I can later add it to the previous estimate $hat{x}_x(k^{-})$, which is also $4x1$.
kalman-filter
I am trying to follow the computation of EKF presented in this paper http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=C9CB210A45F0D7ED5CA7DE174F1A5490?doi=10.1.1.681.8390&rep=rep1&type=pdf (p.16-18), with the only difference that my system state does not include the acceleration, so it is a $4x1$ matrix:
$$x_x(t) = begin{bmatrix}
x(t) \
v_x(t) \
y(t) \
v_y(t)
end{bmatrix} $$
where $x(t)$ and $y(t)$ are the cartesian coordinates and $v_x(t)$ and $v_y(t)$ are the velocity components.
I have problem computing the State Estimate Update step (eq. 17 in the paper), where the measurement residual needs to be calculated. According to the paper, the State Estimate Update is:
$$hat{x}_x(k^{+}) = hat{x}_x(k^{-}) + K(k)[z_{z}(k)-h(hat{x}_x(k^{-}))]$$
where
$$z_z(k) = begin{bmatrix}
z_x(k) \
z_y(k)
end{bmatrix},
$$
where $z_x(k)$ and $z_y(k)$ are the measurements of $x$ and $y$ positions and
$$
h(x_x(k)) = begin{bmatrix}
x(k) &0 &0 &0 \
0 &0 &y(k) &0
end{bmatrix}. $$
$K(k)$ is the Kalman Gain, which in my case is a $4x2$ matrix.
My question is, how do I subtract $h(hat{x}_x(k^{-}))$ which is a $2x4$ matrix from $z_z(k)$, which is a $2x1$ matrix?
If I am not mistaken, the result of $K(k)[z_{z}(k)-h(hat{x}_x(k^{-}))]$ needs to be a $4x1$ matrix so that I can later add it to the previous estimate $hat{x}_x(k^{-})$, which is also $4x1$.
kalman-filter
kalman-filter
asked Nov 15 at 11:23
Anemone
32
32
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1 Answer
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I suspect it is a mistake and the actual equation should have been
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0
end{bmatrix} x(k).
$$
In this case equation $(9)$ would also make more sense. So in your case it would be
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 0 & 1 & 0
end{bmatrix} x(k).
$$
Makes more sense now, thank you!
– Anemone
Nov 16 at 11:14
@Anemone Also equation $(2)$ should not have a dot on top of $x$, so I would not trust everything immediately from that paper directly if I where you, because it might contain more mistakes.
– Kwin van der Veen
Nov 16 at 11:24
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I suspect it is a mistake and the actual equation should have been
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0
end{bmatrix} x(k).
$$
In this case equation $(9)$ would also make more sense. So in your case it would be
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 0 & 1 & 0
end{bmatrix} x(k).
$$
Makes more sense now, thank you!
– Anemone
Nov 16 at 11:14
@Anemone Also equation $(2)$ should not have a dot on top of $x$, so I would not trust everything immediately from that paper directly if I where you, because it might contain more mistakes.
– Kwin van der Veen
Nov 16 at 11:24
add a comment |
up vote
0
down vote
accepted
I suspect it is a mistake and the actual equation should have been
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0
end{bmatrix} x(k).
$$
In this case equation $(9)$ would also make more sense. So in your case it would be
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 0 & 1 & 0
end{bmatrix} x(k).
$$
Makes more sense now, thank you!
– Anemone
Nov 16 at 11:14
@Anemone Also equation $(2)$ should not have a dot on top of $x$, so I would not trust everything immediately from that paper directly if I where you, because it might contain more mistakes.
– Kwin van der Veen
Nov 16 at 11:24
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I suspect it is a mistake and the actual equation should have been
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0
end{bmatrix} x(k).
$$
In this case equation $(9)$ would also make more sense. So in your case it would be
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 0 & 1 & 0
end{bmatrix} x(k).
$$
I suspect it is a mistake and the actual equation should have been
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0
end{bmatrix} x(k).
$$
In this case equation $(9)$ would also make more sense. So in your case it would be
$$
h(x(k)) = begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 0 & 1 & 0
end{bmatrix} x(k).
$$
answered Nov 16 at 9:38
Kwin van der Veen
5,1202826
5,1202826
Makes more sense now, thank you!
– Anemone
Nov 16 at 11:14
@Anemone Also equation $(2)$ should not have a dot on top of $x$, so I would not trust everything immediately from that paper directly if I where you, because it might contain more mistakes.
– Kwin van der Veen
Nov 16 at 11:24
add a comment |
Makes more sense now, thank you!
– Anemone
Nov 16 at 11:14
@Anemone Also equation $(2)$ should not have a dot on top of $x$, so I would not trust everything immediately from that paper directly if I where you, because it might contain more mistakes.
– Kwin van der Veen
Nov 16 at 11:24
Makes more sense now, thank you!
– Anemone
Nov 16 at 11:14
Makes more sense now, thank you!
– Anemone
Nov 16 at 11:14
@Anemone Also equation $(2)$ should not have a dot on top of $x$, so I would not trust everything immediately from that paper directly if I where you, because it might contain more mistakes.
– Kwin van der Veen
Nov 16 at 11:24
@Anemone Also equation $(2)$ should not have a dot on top of $x$, so I would not trust everything immediately from that paper directly if I where you, because it might contain more mistakes.
– Kwin van der Veen
Nov 16 at 11:24
add a comment |
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